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z2+14=9z
We move all terms to the left:
z2+14-(9z)=0
We add all the numbers together, and all the variables
z^2-9z+14=0
a = 1; b = -9; c = +14;
Δ = b2-4ac
Δ = -92-4·1·14
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{25}=5$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-5}{2*1}=\frac{4}{2} =2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+5}{2*1}=\frac{14}{2} =7 $
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